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Nimfa-mama [501]
3 years ago
7

Determine the value of (tangent of 88 degrees, 22 minutes, and 45 seconds). In this case, minutes are 1/60 of a degree and secon

ds are 1/3600 of a degree.a. 23.42583649 c. 35.34015106 b. 56.36842561 d. 47.69869523
Mathematics
1 answer:
mariarad [96]3 years ago
4 0

Answer:

  c.  35.34015106

Step-by-step explanation:

As with many problems of this nature, you only need to get close to be able to choose the correct answer. 22 minutes 45 seconds is just slightly less than 1/2 degree (30 minutes), so the tangent value will be just slightly less than tan(88.5°) ≈ 38. The appropriate choice is 35.34015106.

If you need confirmation, you can find tan(88°) ≈ 29, so you know the answer will be between 29 and 38.

__

The above has to do with strategies for choosing answers on multiple-choice problems. Below, we will work the problem.

The angle is (in degrees) ...

  88 + 22/60 +45/3600 = 88 + (22·60 +45)/3600 = 88 +1365/3600

  ≈ 88.3791666... (repeating) . . . . degrees

A calculator tells you the tangent of that is ...

  tan(88.3791666...°) ≈ 35.3401510614

Many calculators will round that to 10 digits, as in the answer above. Others can give a value correct to 32 digits. Spreadsheet values will often be correct to 15 or 16 digits.

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Answer:

x = 13

Step-by-step explanation:

This question is based on Secant Secant theorem.

Secant Secant theorem gives us the following formula:

(AB + BD)AB = (AC + CE).AC

From the above question we have the following parameters

AB = 5

BD = x

AC = 7.5

CE = 4.5

Hence,

(AB + BD)AB = (AC + CE).AC

(5 + x)5 = (7.5 + 4.5)7.5

25 + 5x = 90

Collect like terms

5x = 90 - 25

5x = 65

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A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first the
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Answer:

The hotel charge in each city before tax was <u>$5125</u> of the first city and <u>$3625</u> of the second city.

Step-by-step explanation:

Given:

A theater group made appearances into cities the hotel charge before tax and the second city was 1500 higher than the first.

The tax of the first city was 6% and the tax of the second city was 10%.

Total hotel tax paid for two cities with $670.

Now, to find the hotel charge in each city before tax.

Let the hotel charge in first city before tax be x.

And the hotel charge in second city before tax be y.

<em>So, as the hotel charge of the second city was 1500 higher than the first.</em>

<em>Thus</em>,

y=x-1500   ........(1)

<em>And as given, the tax of the first city was 6% and the tax of the second city was 10%, total hotel tax paid for two cities with $670.</em>

6% of x + 10% of y = $670.

\frac{6x}{100} +\frac{10y}{100} =670

0.06x+0.10y=670

Substituting the value of y from equation (1) we get:

0.06x+0.10(x-1500)=670

0.06x+0.10x-150=670

0.16x-150=670

<em>Adding both sides by 150 we get:</em>

0.16x=820

<em>Dividing both sides by 0.16 we get:</em>

x=5125.

<em>The hotel charge in first city before tax = $5125.</em>

Now, substituting the value of x in equation (1) we get:

y=x-1500

y=5125-1500

y=3625.

<em>The hotel charge in second city before tax = $3625.</em>

Therefore, the hotel charge in each city before tax was $5125 of the first city and $3625 of the second city.

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